ГДЗ по алгебре 9 класс Макарычев Задание 92

\[\boxed{\text{92\ (92).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[y_{1}(0,5) = {0,5}^{2} = 0,25\]

\[y_{2}(0,5) = 1,8 \cdot {0,5}^{2} =\]

\[= 1,8 \cdot 0,25 = 0,45\]

\[y_{3}(0,5) = \frac{1}{3} \cdot {0,5}^{2} =\]

\[= \frac{1}{3} \cdot 0,25 \approx 0,083 \Longrightarrow\]

\[\Longrightarrow y_{3} < y_{1} < y_{2}.\]

\[y_{1}(1) = 1^{2} = 1\]

\[y_{2}(1) = 1,8 \cdot 1^{2} = 1,8\]

\[y_{3}(1) = \frac{1}{3} \cdot 1^{2} \approx 0,3 \Longrightarrow\]

\[\Longrightarrow y_{3} < y_{1} < y_{2}.\]

\[y_{1}(2) = 2^{2} = 4\]

\[y_{2}(2) = 1,8 \cdot 2^{2} = 1,8 \cdot 4 = 7,2\]

\[y_{3}(2) = \frac{1}{3} \cdot 2^{2} = \frac{4}{3} \approx 1,3\]

\[\Longrightarrow y_{3} < y_{1} < y_{2}.\]